# Question

Formatted question description: https://leetcode.ca/all/272.html

Given the root of a binary search tree, a target value, and an integer k, return the k values in the BST that are closest to the target. You may return the answer in any order.

You are guaranteed to have only one unique set of k values in the BST that are closest to the target.

Example 1: Input: root = [4,2,5,1,3], target = 3.714286, k = 2
Output: [4,3]


Example 2:

Input: root = , target = 0.000000, k = 1
Output: 


Constraints:

• The number of nodes in the tree is n.
• 1 <= k <= n <= 104.
• 0 <= Node.val <= 109
• -109 <= target <= 109

Follow up: Assume that the BST is balanced. Could you solve it in less than O(n) runtime (where n = total nodes)?

# Algorithm

This tree is considered balanced if: the difference between heights of the left subtree and right subtree is not more than 1.

### In-order

The comparison is completed during the in-order traversal.

When traversing to a node,

• If the result array is less than k at this time, add this node value directly to the result,
• If the absolute value of the difference between the node value and the target value is less than the absolute value of the difference between the first element of the result and the target value, indicating that the current value is closer to the target value, delete the first element and add the current node value to the end,
• On the contrary, it means that the current value deviates more from the target value than all the values in the result res. Due to the characteristics of the in-order traversal, the subsequent values will deviate further, so the final result is directly returned at this time

### Heap

A pair of difference diff and node value stored in the heap.

In order to traverse the binary tree (other traversal methods can also be used), and then calculate the absolute value of the difference between the target value and the target value for each node value.

Due to the nature of the maximum heap, the largest diff is automatically the first to maintain k pairs, if If there are more than k, delete the big pair at the front of the heap, and remove the node value in the pair for the k pairs left and store it in the result.

# Code

• import java.util.LinkedList;
import java.util.List;

public class Closest_Binary_Search_Tree_Value_II {

public static void main(String[] args) {
Closest_Binary_Search_Tree_Value_II out = new Closest_Binary_Search_Tree_Value_II();
Solution s = out.new Solution();

TreeNode root = new TreeNode(4);
root.left = new TreeNode(2);
root.right = new TreeNode(5);

root.left.left = new TreeNode(1);
root.left.right = new TreeNode(3);

System.out.println(s.closestKValues(root, 3.714286, 2));
}

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/

public class Solution {

public List<Integer> closestKValues(TreeNode root, double target, int k) {
inOrderTraversal(root, target, k);
return result;
}

private void inOrderTraversal(TreeNode root, double target, int k) {
if (root == null) {
return;
}
inOrderTraversal(root.left, target, k);
if (result.size() < k) {
} else if(result.size() == k) { // hand-made priority queue
if (Math.abs(result.getFirst() - target) > (Math.abs(root.val - target))) {
result.removeFirst();
} else { // meaning, current node value is further than both first and last value of result list
return; // diff is larger, so skip, as trim
}
}
inOrderTraversal(root.right, target, k);
}
}
}

############

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private List<Integer> ans;
private double target;
private int k;

public List<Integer> closestKValues(TreeNode root, double target, int k) {
this.target = target;
this.k = k;
dfs(root);
return ans;
}

private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
if (ans.size() < k) {
} else {
if (Math.abs(root.val - target) >= Math.abs(ans.get(0) - target)) {
return;
}
ans.remove(0);
}
dfs(root.right);
}
}

• // solution with heap and diff
class Solution {
public:
vector<int> closestKValues(TreeNode* root, double target, int k) {
vector<int> res;
priority_queue<pair<double, int>> q;
inorder(root, target, k, q);
while (!q.empty()) {
res.push_back(q.top().second);
q.pop();
}
return res;
}
void inorder(TreeNode *root, double target, int k, priority_queue<pair<double, int>> &q) {
if (!root) return;
inorder(root->left, target, k, q);
q.push({abs(root->val - target), root->val});
if (q.size() > k) q.pop();
inorder(root->right, target, k, q);
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def closestKValues(self, root: TreeNode, target: float, k: int) -> List[int]:
def dfs(root):
if root is None:
return
dfs(root.left)
if len(q) < k:
q.append(root.val)
else:
if abs(root.val - target) >= abs(q - target):
return # trim: meaning, current node value is further than both first and last value of result list
q.popleft()
q.append(root.val)
dfs(root.right)

q = deque()
dfs(root)
return list(q)

from collections import deque

class Solution: # iterative
def closestKValues(self, root, target, k):
result = []
stack = []
while root or stack:
while root:
stack.append(root)
root = root.left

root = stack.pop()
if len(result) < k:
result.append(root.val)
else:
if abs(root.val - target) < abs(result - target):
result.pop(0)
result.append(root.val)
else:
break

root = root.right

return result

• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func closestKValues(root *TreeNode, target float64, k int) []int {
var ans []int
var dfs func(root *TreeNode)
dfs = func(root *TreeNode) {
if root == nil {
return
}
dfs(root.Left)
if len(ans) < k {
ans = append(ans, root.Val)
} else {
if math.Abs(float64(root.Val)-target) >= math.Abs(float64(ans)-target) {
return
}
ans = ans[1:]
ans = append(ans, root.Val)
}
dfs(root.Right)
}
dfs(root)
return ans
}